Edexcel Jun 2015 (IAL) Paper 5 Q21 (with explained solutions) |

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Edexcel Jun 2015 (IAL) Paper 5 Q21

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SECTION B Answer ALL the questions. Write your answers in the spaces provided. 21 Steel is the worlds most important structural metal; it is strong and cheap but it corrodes rapidly if unprotected. In its simplest form, steel is an alloy of iron and carbon. (a) The following method was used to determine the percentage of iron in a sample of wire. Exactly 1.25 g of the wire was placed in a conical flask and about 50 cm3 of dilute sulfuric acid (an excess) was added. The flask was closed as shown in the diagram below. glass rod glass tube rubber tube with slit When all of the iron in the wire had been converted to iron(II) sulfate, the contents of the flask were used to make 250.0 cm3 of solution with distilled water. 25.00 cm3 portions of this final solution were placed in a conical flask, acidified with an equal volume of dilute sulfuric acid and then titrated with a potassium manganate(VII) solution of concentration 0.0195 mol dm3. The mean titre was 22.15 cm3. (i) Write the equation for the reaction between iron and dilute sulfuric acid. Include state symbols in your answer. (1) *P45074A01128* 11 Turn over<br />
(ii) Suggest why the conical flask was not left open, and how the labelled part of the apparatus shown in the diagram works. (2)(iii) State the essential steps of the procedure for making up the reaction mixture to 250.0 cm3 for use in the titration. (3)(iv) Write the ionic equation for the titration reaction to show that 5 mol of iron(II) ions react with 1 mol of manganate(VII) ions. State symbols are not required. (1) 12 *P45074A01228*<br />
(v) Calculate the percentage by mass of iron in the wire. Give your answer to three significant figures. (4) (vi) Describe the colour change at the end-point of the titration. (1)*P45074A01328* 13 Turn over<br />
(vii) One student who carried out this experiment forgot to acidify the mixture in the conical flask before the titration. A brown precipitate formed before the end-point. Identify the brown precipitate and explain how this error affects the titration value. (3)(b) The rusting of iron is an electrochemical process. A piece of iron with a droplet of water on its surface operates as an electrochemical cell. In the first stage of corrosion, iron is oxidized to iron(II) ions in the anodic area and the electrons produced travel to the cathodic area where oxygen from the air is reduced. iron water anodic area cathodic area 14 *P45074A01428*<br />
(i) From the information about standard electrode potentials on pages 14 and 15 of the Data Booklet, write the ionic half equations for the reactions taking place at the anodic area and at the cathodic area. State symbols are not required. Anodic area Cathodic area (ii) Calculate E (cid:57) cell for the overall reaction in (b)(i). (iii) By considering the rusting mechanism described in part (b), suggest why the presence of salt in the water droplet speeds up rusting. (2) (1) (1)(iv) The corrosion of oil pipelines made of steel is prevented by connecting the pipeline to magnesium blocks. Suggest how this method works. (1)(Total for Question 21 = 20 marks) *P45074A01528* 15 Turn over<br />

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Section B Question Number 21(a)(i) Fe(s) + H2SO4(aq) FeSO4(aq) + H2(g) Acceptable Answer OR Fe(s) + 2H+(aq) Fe2+(aq) + H2(g) OR ionic equations including sulfate ions OR multiples Acceptable Answer Question Number 21(a)(ii) Otherwise the Fe2+ formed will oxidize ALLOW So air / oxygen cannot enter the flask To prevent reaction with air /oxygen (1) Hydrogen can escape through the slit OR So pressure does not build up (1) IGNORE Acid spray Question Number 21(a)(iii) Transfer the reaction mixture to a Acceptable Answer (250 cm3) volumetric/graduated flask ALLOW standard flask (1) (Rinse conical flask and) add washings to the volumetric flask (1) Make solution up to the mark (with distilled water/sulfuric acid) and then mix ALLOW any indication of mixing (1) IGNORE Filtration Reject Mark Mark Reject Iron/steel oxidized Reject Mark Using other liquids WCH0501msc20150812.docx 11 July 2015<br />
Question Number 21(a)(iv) 5Fe2+ + MnO4+ 8H+ Acceptable Answer5Fe3+ + Mn2+ + 4H2O OR multiples Ignore state symbols even if incorrect Question Number 21(a)(v) Amount MnO4 Acceptable Answer= 22.15 x 0.0195 /1000 (1) = 4.31925 x 104 ans* Amount Fe2+ = 5 x ans* (1) = 2.159625 x 103 ans** Mass of iron in wire = 10 x ** x 55.8 (1) = 1.20507 (g) ans*** % purity = 100 x ans*** / 1.25 = 96.40566 = 96.4 % (1) Ignore rounding errors until final answer Correct answer (96.4%) with or without working scores 4 ALLOW Use of Ar(Fe) = 56 when= 22.15 x 0.0195 /1000 Amount MnO4 (1) = 4.31925 x 104 ans* Amount Fe2+ = 5 x ans* (1) = 2.159625 x 103 ans** Mass of iron in wire = 10 x ** x 56 (1) Mass of iron in wire = 1.20939 % purity = 96.7512 = 96.8 % (1) Ignore intermediate rounding until final answer Correct answer (96.8%) with or without working scores 4 TE on each stage in the calculation % purity > 100 scores max 2 Reject Mark Reject Mark Answer not to 3 SF Answer not to 3 SF WCH0501msc20150812.docx 11 July 2015<br />
Question Number 21(a)(vi) Colourless / pale yellow Acceptable Answer to (pale) pink / first permanent pink Question Number 21(a)(vii) (More manganate(VII) is needed to Acceptable Answer Mark Mark Reject Purple Just (pale) pink Reject Mn(OH)2 Reject Mark Question Number 21(b)(i) Anodic area: oxidize Fe2+, so) titre will be larger (1) Stand alone mark Because the Mn oxidation number changes from 7 to 4 (rather than 2) OR Mn accepts fewer electrons per mole (1) (Brown precipitate is) manganese(IV) oxide / MnO2 ALLOW Mn(OH)4 IGNORE References to inaccurate / inconsistent titre values (1) OR Acceptable Answer Fe2+ + 2e()Fe (Eo = 0.44 V) FeFe2+ + 2e() (1) O2 + 2H2O + 4e()4OH (Eo = +0.40 V) O2 + 2H+ + 2e()H2O (Eo = +1.23 V) (1) ALLOW Cathodic area: Penalise omission of electrons or use of cell diagrams once only Anode and cathode reversed max 1. IGNORE State symbols even if incorrect Single arrow in equations WCH0501msc20150812.docx 11 July 2015<br />
Reject Mark Question Number 21(b)(ii) Eo Acceptable Answer cell = (+)0.40(0.44) = cell = (+)1.23(0.44) = (+)0.84 (V) ALLOW Eo (+)1.67 (V) Correct answer with no working scores 1 Question Number 21(b)(iii) Dissolved salt makes the water Acceptable Answer a better conductor (of ions) OR The solution acts like a salt bridge OR Makes it an (effective) electrolyte OR Improves the flow of ions through the solution ALLOW Improves the flow of electrons through the metal Reject Improves the flow of ions through the metal Improves the flow of electrons through the solution Mark Mark Question Number 21(b)(iv) Magnesium has a more negative Eo Acceptable Answer (allow more reactive) and so reduces the Fe2+ OR suppresses the oxidation of iron OR forces the iron (in the absence of oxygen) to act as the cathode ALLOW Mg corrodes / oxidizes in preference to / faster than (the Fe / steel) OR Magnesium acts as a sacrificial anode Reject Just sacrificial protection Total for Question 21 = 20 marks WCH0501msc20150812.docx 11 July 2015<br />

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